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I have this circuit:

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On start-up, the power supply experiences an overcurrent condition, because when the capacitor is charged, it looks like a short circuit as seen from the power supply. How can I prevent this?

Adding a current-limiting resistor before the capacitor is not suitable for me, because of the load-dependent voltage drop - the load, of course, will look smaller once the capacitor has charged at the maximum current of about 2 A. Maybe there is a simple circuit to temporarily enable a current-limiting resistor? Or another simple solution?

ADDED. I have these ideas. Please feel free to comment.

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  • \$\begingroup\$ Your "added" resistor in series with the capacitor will reduce the capacitor's ability to do useful filtering. This is because the load would have to draw the capacitor's current through that same resistor. \$\endgroup\$
    – gbarry
    Jan 11, 2013 at 18:40
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    \$\begingroup\$ A correction to your terminology: the impedance of the capacitor does not change. You are right that at start there will be a high inrush current. However, impedance is a combination of resistance, inductance, and capacitance, so in this case the impedance of the capacitor is constant, it's just the capacitance, plus whatever non-ideal (and often negligible) resistance and inductance it has. \$\endgroup\$
    – Phil Frost
    Jan 11, 2013 at 20:14

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Well-designed power supplies get around capacitor load charging in a few ways:

  • a pre-charge, mate-first contact that has a resistor or NTC thermistor in series with the contact; this path pre-charges the cap through the resistance and gets shorted out when the other power pins mate

  • a soft-start feature where the output voltage ramps up over tens or hundreds of milliseconds, decreasing the capacitor charging current

  • a brick-wall or constant-current mode of operation; when there's heavy load, the power supply goes into current regulation mode (it keeps the output current constant and lets the voltage drop out) which will charge any capacitors in a controlled manner

Since your power supply doesn't appear to have any of these features, you need an external solution. Several manufacturers make what are called hot-swap controllers which essentially act like programmable resistors (using MOSFETs) to apply a load to a power supply output in a controlled manner, independent of the power supply itself.

This allows for gradual charging of capacitors without overwhelming the power supply, plus once the caps are charged, the low \$R_{ds(on)}\$ of the MOSFET makes the losses manageable.

Most of the time, there's no need for a series current limiting resistor when using a filter capacitor. That being said, some types of capacitors (tantalum and some organic semiconductor types) do need charge/discharge current limiting.

You haven't mentioned the wattage of your power supply or how much capacitor you're trying to charge - this information is obviously needed to judge whether or not the cost and complexity of a hot-swap controller is justified.

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I think what you are looking for is called an "inrush current limiter". Here's a search on Digi-key that you might find what you are looking for. And here is a Wikipedia article on the subject.

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If your ciruit has a "reset chip", and if it won't draw much current while in reset, I'd suggest perhaps using a MOSFET to turn input current on and off, but with a moderate-value resistor to allow some current through while the MOSFET is off. Until the capacitor has charged up to the reset threshold and been above that point long enough for the reset chip to be happy (meaning it will have had time to charger further above the threshold), the current into the device will be limited but the device itself won't use much current. Once the reset is happy, the MOSFET will provide an excellent conduction path, thus allowing the device to make good use of the supply current.

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A solution from the old days was to put a resistor in series, and then use a relay to short-circuit the resistor. The relay coil goes across the capacitor, so that the relay pulls in after the voltage has risen to a suitable level.

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  • \$\begingroup\$ This is bad solution for me. Relay is big and expensive. Is there simple way to change the relay with transistor? \$\endgroup\$
    – user54579
    Jan 11, 2013 at 15:42
  • \$\begingroup\$ I didn't say you had to do it that way :) Yes, you can use a transistor; that's what all these other "current limiting" devices are about. But it's not as simple as the relay. I'm sure someone will post a circuit shortly. \$\endgroup\$
    – gbarry
    Jan 11, 2013 at 15:48
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Another solution is to specify a DC power supply with current limiting (say at 3A, if your load is 2A) instead of tripping when it senses overcurrent.

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  • \$\begingroup\$ It's suitable only if the power supply can change its current-limiting value. My is not. \$\endgroup\$
    – user54579
    Jan 11, 2013 at 15:46
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Usually, you try to make sure that your capacitor is smaller than the maximum output capacitance allowed for your supply. Some DC-DC converters allow as little as 100 µF. If the output characteristic of your supply is of the fold-back type, this is a very important piece of information - and you would need a supply that doesn't shut down or hiccup during over-current events if you want to use any size of capacitor. If there is no such rating in the data sheet of your supply, you are relying on pure luck, even if you are experimenting with series resistors.

The questions it always boils down to: Will your power supply go into shut-down, and how fast will it go there? If it shuts down faster than it takes you to charge the capacitor and obtain a low current, it won't work.

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